Write the linear equation in the slope-intercept form when m=1 and b=-12

9+9-4-123456-5431-87+0-6=5 x 3<br> answer please!!!!1

Serjik [45]

Answer:

-177,856 does not equal 15

-177,856=15

3 0

2 years ago

What value of n makes the equation true?

dybincka [34]

1(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20 i say the answer 1 or 10

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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10

30

4 0

1 year ago

Write each expression as an algebraic (nontrigonometric) expression in u, u > 0.

max2010maxim [7]

Answer:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

Step-by-step explanation:

We want to write the trignometric expression:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)\text{ where } u>0$

As an algebraic equation.

First, we can focus on the inner expression. Let θ equal the expression:

$\displaystyle \theta=\sec^{-1}\left(\frac{u}{10}\right)$

Take the secant of both sides:

$\displaystyle \sec(\theta)=\frac{u}{10}$

Since secant is the ratio of the hypotenuse side to the adjacent side, this means that the opposite side is:

$\displaystyle o=\sqrt{u^2-10^2}=\sqrt{u^2-100}$

By substitutition:

$\displaystyle= \sin(2\theta)$

Using an double-angle identity:

$=2\sin(\theta)\cos(\theta)$

We know that the opposite side is √(u² -100), the adjacent side is 10, and the hypotenuse is u. Therefore:

$\displaystyle =2\left(\frac{\sqrt{u^2-100}}{u}\right)\left(\frac{10}{u}\right)$

Simplify. Therefore:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

4 0

2 years ago

4. What’s the answer to this question?!

telo118 [61]

Answer:

2000

Step-by-step explanation:

3 0

3 years ago

Hello there I need help.

andriy [413]

Answer:

A. 40

Step-by-step explanation:

Q1 - 40

Q2 - 65

Q3 - 80

And 90 is max

So 40 is the correct answer.

5 0

2 years ago